Primitive element theorem - Imported Knowlpedia Content

Let $L/K$ be a finite field extension.

Theorem (Primitive element theorem). If $L/K$ is a finite separable extension, then there exists an element $\alpha\in L$ such that

L = K(\alpha),

i.e. $L/K$ is a simple extension.

Remarks

A useful strengthening (often used in proofs) is: if $L=K(\beta,\gamma)$ is finite and separable and $K$ is infinite, then for all but finitely many $c\in K$ , the element $\alpha=\beta+c\gamma$ satisfies $L=K(\alpha)$ .

Examples

$\mathbb{Q}(\sqrt2,\sqrt3)/\mathbb{Q}$ is separable because $\operatorname{char}(\mathbb{Q})=0$ (see characteristic 0 or prime). A primitive element is $\alpha=\sqrt2+\sqrt3$ . One checks $\mathbb{Q}(\alpha)$ contains both $\sqrt2$ and $\sqrt3$ , hence equals $\mathbb{Q}(\sqrt2,\sqrt3)$ .

Any finite field extension $\mathbb{F}_{p^n}/\mathbb{F}_p$ is separable (finite fields are perfect). Therefore $\mathbb{F}_{p^n}=\mathbb{F}_p(\alpha)$ for some $\alpha$ . Concretely, choosing $\alpha$ to generate the cyclic group $\mathbb{F}_{p^n}^\times$ from cyclicity of the multiplicative group guarantees $\mathbb{F}_p(\alpha)=\mathbb{F}_{p^n}$ .

If $f\in K[x]$ is separable and $L$ is its splitting field over $K$ , then $L/K$ is finite and separable; hence $L=K(\alpha)$ for some $\alpha\in L$ . For instance, the splitting field of $x^3-2$ over $\mathbb{Q}$ can be generated by a single element $\alpha\in \mathbb{Q}(\sqrt[3]{2},\zeta_3)$ .

(For contrast: without separability, the conclusion can fail; there exist finite inseparable extensions that are not simple.)